Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{k^3 + 12k^2 + 27k}{6k^3 + 102k^2 + 432k}$
First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {k(k^2 + 12k + 27)} {6k(k^2 + 17k + 72)} $ $ r = \dfrac{k}{6k} \cdot \dfrac{k^2 + 12k + 27}{k^2 + 17k + 72} $ Simplify: $ r = \dfrac{1}{6} \cdot \dfrac{k^2 + 12k + 27}{k^2 + 17k + 72}$ Since we are dividing by $k$ , we must remember that $k \neq 0$ Next factor the numerator and denominator. $ r = \dfrac{1}{6} \cdot \dfrac{(k + 9)(k + 3)}{(k + 9)(k + 8)}$ Assuming $k \neq -9$ , we can cancel the $k + 9$ $ r = \dfrac{1}{6} \cdot \dfrac{k + 3}{k + 8}$ Therefore: $ r = \dfrac{ k + 3 }{ 6(k + 8)}$, $k \neq -9$, $k \neq 0$